Let $I\subset S$ be a $\mathbf{t}$-spread strongly stable ideal such that $I\subseteq {\mathfrak{m}}^2$. Then I is Cohen-Macaulay if and only if there exists $\ell \le d$ such that

Firstly, we prove that I is vertex splittable. If I is a principal ideal, then it is vertex splittable. Suppose now that $\left|\mathcal{G}\right(I\left)\right|>1$. We can write $I=x_1{I}_1+I_2$, where $\mathcal{G}\left(I_1\right)=\{u/x_1:u\in \mathcal{G}\left(I\right),x_{1}dividesu\}$ and $\mathcal{G}\left(I_2\right)=\mathcal{G}\left(I\right)\setminus \mathcal{G}\left(x_1{I}_1\right)$. It is immediate to see that $I_1\subset S$ is $(t_2,\cdots ,t_{d-1})$-spread strongly stable and that $I_2$ is a $\mathbf{t}$-spread strongly stable ideal of $K[x_2,\cdots ,x_n]$. By induction on n and on the highest degree of a generator of I, we have that $I_1$ and $I_2$ are vertex splittable. Hence, so is I.

We may suppose that $x_n$ divides some minimal generator of I. Otherwise, we can consider I as a monomial ideal of a smaller polynomial ring. If I is principal, then we have $I=\left(u\right)=(x_1{x}_{1+t_1}\cdots x_{1+t_1+\cdots +t_{\ell -1}})$, with $n=1+t_1+\cdots +t_{\ell -1}$, and $\ell \le d$. Otherwise, if I is not principal, then $pdS/I>1$ and we can write $I=x_1{I}_1+I_2$ as above. By Theorem 1, $I_2$ is Cohen-Macaulay and $pdS/I_2=pdS/I+1>1$. Thus $I_2\ne 0$. Hence, by induction there exists $\ell \le d$ such that

Since $\mathcal{G}\left(I_2\right)\subset \mathcal{G}\left(I\right)$, the assertion follows. □

We may assume that all variables $x_i$ divide some minimal monomial generator of I. It is noted in the proof of [10] that for any variable $x_i$ which divides a minimal monomial generator of minimal degree of I, we can write $I=x_i{I}_1+I_2$, where $\mathcal{G}\left(x_i{I}_1\right)=\{u\in \mathcal{G}\left(I\right):x_{i}dividesu\}$, $\mathcal{G}\left(I_2\right)=\mathcal{G}\left(I\right)\setminus \mathcal{G}\left(x_i{I}_1\right)$ and the following properties are satisfied:

By induction on n and on the highest degree of a generator of I, it follows that both $I_1$ and $I_2$ are vertex splittable. Hence, so is I. □

[11]Let $I\subset S$ be a polymatroidal ideal generated in degree $d\ge 2$ and let $x_i$ be a variable dividing some monomial of $\mathcal{G}\left(I\right)$. Then $(I:x_i)$ is polymatroidal. If, in addition, I is Cohen-Macaulay, then $(I:x_i)$ is also Cohen-Macaulay.

Let $n,d\ge 1$ be positive integers. Then

Since $dimS/{\mathfrak{m}}^d=0$ we have $depthS/{\mathfrak{m}}^d=0$ and $pdS/{\mathfrak{m}}^d=n$. If $d=1$, then $I_{n,1}=\mathfrak{m}$ and $pdS/I_{n,1}=n$. If $1<d\le n$, we notice that $I_{n,d}=x_n{I}_{n-1,d-1}+I_{n-1,d}$ is a vertex splitting. By formula (1) and induction on n and d, as wanted □

Let I be a polymatroidal ideal. If I is principal, then it is Cohen-Macaulay. Now, let $\left|\mathcal{G}\right(I\left)\right|>1$. If $I=\mathfrak{m}$, then I is Cohen-Macaulay. Thus, we assume that I is generated in degree $d\ge 2$, and that all variables $x_i$ divide some minimal monomial generator of I. By Proposition 1 and Corollaries 5, 5, we have a vertex splitting $I=x_i{I}_1+I_2$ for each variable $x_i$, and $I_1$ and $I_2$ are Cohen-Macaulay polymatroidal ideals with $depthS/I_1=depthS/I_2-1$. Thus $pdS/I_1=pdS/I_2+1$. We may assume that $x_i=x_n$.

By induction on n and on d, it follows that $I_1$ is either a principal ideal, a Veronese ideal, or a squarefree Veronese ideal, and the same possibilities occur for $I_2$. We distinguish the various possibilities.

Case 1. Let $I_2$ be a principal ideal, then $pdS/I_2=1$. Thus $pdS/I_1=2$.

Under this assumption, $I_1$ cannot be principal because $pdS/I_1=2\ge 1$.

Assume that $I_1$ is a Veronese ideal in m variables. Since all variables of S divide some monomial of $\mathcal{G}\left(I\right)$ and $I_2\subset I_1$, then $I_1={\mathfrak{m}}^d$ or $I_1={\mathfrak{n}}^d$, where $\mathfrak{m}=(x_1,\cdots ,x_n)$ and $\mathfrak{n}=(x_1,\cdots ,x_{n-1})$. Thus $m=n$ or $m=n-1$. Lemma 1 implies $m=pdS/I_1=2$. So, $n=2$ or $n=3$.

If $n=2$, then $I=x_2{(x_1,x_2)}^{d-1}+\left(x_1^d\right)={(x_1,x_2)}^d$ which is Cohen-Macaulay and Veronese, or $I=x_2\left(x_1^{d-1}\right)+\left(x_1^d\right)=x_1^{d-1}(x_1,x_2)$ which is not Cohen-Macaulay.

Otherwise, if $n=3$ then $I=x_3{(x_1,x_2,x_3)}^{d-1}+\left(u\right)$ or else $I=x_3{(x_1,x_2)}^{d-1}+\left(u\right)$ where $u\in K[x_1,x_2]$ is a monomial of degree d. If $d=2$, then one easily sees that only in the second case and for $u=x_1{x}_2$ we have that I a Cohen-Macaulay polymatroidal ideal, which is the squarefree Veronese $I_{3,2}$. Otherwise, suppose $d\ge 3$. We may assume that $x_1^2$ divides u. In the first case, $(I:x_1)=x_3{(x_1,x_2,x_3)}^{d-2}+(u/x_1)$ is not principal, neither Veronese, neither squarefree Veronese. Thus, by induction $(I:x_1)$ is not a Cohen-Macaulay polymatroidal ideal, and by Corollary 5 we deduce that I is also not Cohen-Macaulay. Similarly, in the second case, we see that $(I:x_1)=x_3{(x_1,x_2)}^{d-1}+(u/x_1)$, and thus also I, is not a Cohen-Macaulay polymatroidal ideal.

Assume now that $I_1$ is a squarefree Veronese ideal in m variables. Then as argued in the case 1.2 we have $m=n$ or $m=n-1$. Lemma 1 gives $pdS/I_1=m+1-(d-1)=2$. Thus $d=m$. Hence $d=n$ or $d=n-1$. So $I=x_n{I}_{n,n-1}+\left(u\right)$ or $I=x_n{I}_{n-1,n-2}+\left(u\right)$ where $u\in K[x_1,\cdots ,x_{n-1}]$ is a monomial of degree $d=n$ in the first case or $d=n-1$ in the second case. In the first case there is i such that $x_i^2$ divides u. Say $i=1$. Then $(I:x_1)=x_n(I_{n,n-1}:x_1)+(u/x_1)$ is not a principal ideal, neither a Veronese ideal, neither a squarefree Veronese. Therefore, by induction $(I:x_1)$ we see that is not a Cohen-Macaulay polymatroidal ideal, and by Corollary 5I is also not a Cohen-Macaulay polymatroidal ideal. Similarly, in the second case, if $u=x_1\cdots x_{n-1}$ then $I=I_{n,n-1}$ is a Cohen-Macaulay squarefree Veronese ideal. Otherwise $x_i^2$, say with $i=1$, divides u and then, arguing as before, we see that I is not a Cohen-Macaulay polymatroidal ideal.

Case 2. Let $I_2$ be a Veronese ideal in m variables, then $pdS/I_2=m\le n-1$.

Under this assumption, $I_1$ cannot be principal because $pdS/I_1=m+1>1$.

Assume that $I_1$ is a Veronese ideal in ℓ variables. Then $\ell =n$ or $\ell =n-1$. Lemma 1 implies that $\ell =pdS/I_1=pdS/I_2+1=m+1$. Thus $\ell =n$ and $m=n-1$ or $\ell =n-1$ and $m=n-2$. In the first case $I=x_n{\mathfrak{m}}^{d-1}+{(x_1,\cdots ,x_{n-1})}^d={\mathfrak{m}}^d$ is a Cohen-Macaulay Veronese ideal. In the second case, up to relabeling we can write $I=x_n{(x_1,\cdots ,x_{n-1})}^{d-1}+{(x_1,\cdots ,x_{n-2})}^d$. But this ideal is not polymatroidal. Otherwise, by the exchange property applied to $u=x_n{x}_{n-1}^{d-1}$ and $v=x_{n-2}^d$, we should have $x_{n-2}{x}_{n-1}^{d-1}\in I$, which is not the case.

Assume now that $I_1$ is a squarefree Veronese ideal in ℓ variables.Then $\ell =n$ or $\ell =n-1$. Lemma 1 implies that $pdS/I_1=\ell +1-(d-1)=\ell +2-d=pdS/I_2+1=m+1$. Thus, either $m=n+1-d$ or $m=n-d$. Up to relabeling, we have either $I=x_n{I}_{n,d-1}+{(x_1,\cdots ,x_{n+1-d})}^d$ or $I=x_n{I}_{n-1,d-1}+{(x_1,\cdots ,x_{n-d})}^d$. If $d=2$, then these ideals become either $I={(x_1,\cdots ,x_n)}^2$ which is a Cohen-Macaulay Veronese ideal, or $I=x_n(x_1,\cdots ,x_{n-1})+{(x_1,\cdots ,x_{n-2})}^2$ which is not polymatroidal because the exchange property does not hold for $u=x_n{x}_{n-1}$ and $v=x_{n-2}^2$ since $x_{n-1}{x}_{n-2}\notin I$. If $d\ge 3$, then the above ideals are not polymatroidal. In the first case the exchange property does not hold for $u=x_{n+1-d}^d\in I_2$ and $v=(x_{n+2-d}\cdots x_n)x_n\in x_n{I}_1$, otherwise $x_j{x}_{n+1-d}^{d-1}\in I$ for some $n+2-d\le j\le n$, which is not the case.

Case 3. Let $I_2$ be a squarefree Veronese in m variables, $m\le n-1$. Then Lemma 1 implies $pdS/I_2=m+1-d$ with $d\le m$. Hence $pdS/I_1=m+2-d$.

In such a case the ideal $I_1$ cannot be principal because $pdS/I_1=m+2-d>1$.

Assume that $I_1$ is a Veronese ideal in ℓ variables. Then $\ell =n$ or $\ell =n-1$. Lemma 1 implies that $\ell =pdS/I_1=m+2-d$. Hence either $d=m-n+2$ or $d=m-n+3$. Since $m\le n-1$, either $d\le 1$ or $d\le 2$. Only the case $d=2$ is possible. If $d=2$, then $\ell =m=n-1$ and we have $I=x_n(x_1,\cdots ,x_{n-1})+{(x_1,\cdots ,x_{n-1})}^2$. This ideal is not Cohen-Macaulay, otherwise it would be height-unmixed. Indeed $(I:x_n)=(x_1,\cdots ,x_{n-1})$ and $(I:x_{n-1})=(x_1,\cdots ,x_n)$ are two associated primes of I having different height.

Finally, assume that $I_1$ is a squarefree Veronese ideal in ℓ variables.Then $\ell =n$ or $\ell =n-1$. Lemma 1 implies that $pdS/I_1=\ell +1-(d-1)=\ell +2-d=m+2-d$. Thus $\ell =m$ and so either $\ell =m=n$ or $\ell =m=n-1$. The first case is impossible because $m\le n-1$. In the second case, we have $I=x_n{I}_{n-1,d-1}+I_{n-1,d}=I_{n,d}$ which is a Cohen-Macaulay squarefree Veronese ideal. □

Theorem 2. (Dirac, [7]).A finite simple graph G is chordal if and only if G admits a perfect elimination order.

Theorem 3. (Fröberg, [13]).Let G be a finite simple graph. Then, $I\left(G\right)$ has a linear resolution if and only if G is cochordal.

Let G be a finite simple graph. Then $I\left(G\right)$ has linear resolution if and only if $I\left(G\right)$ is vertex splittable. Furthermore, if $x_1>\cdots >x_n$ is a perfect elimination order of $G^c$, then

In particular, if any of the equivalent conditions hold, then

We proceed by induction on $n=\left|V\right(G\left)\right|$. By Theorem 3, G must be cochordal. Fix $x_1>\cdots >x_n$ a perfect elimination order of $G^c$. By Proposition 2, $I\left(G\right)=x_1(x_j:j\in N_G\left(1\right))+I(G\setminus \left\{1\right\})$ is a vertex splitting. Applying Theorem 1, $I\left(G\right)$ is Cohen-Macaulay if and only if $J=(x_j:j\in N_G\left(1\right))$ and $I(G\setminus \{1\left\}\right)$ are Cohen-Macaulay and $pdS/I\left(G\right)=pdS/J=pdS/I(G\setminus \{1\left\}\right)+1$. J is Cohen-Macaulay because it is an ideal generated by variables and $pdS/J=|N_G\left(1\right)|$. Notice that $x_2>\cdots >x_n$ is a perfect elimination order of ${(G\setminus \left\{1\right\})}^c$.

If $I(G\setminus \{1\left\}\right)=0$, then $pdS/I\left(G\right)=pdS/J=pdS/I(G\setminus \{1\left\}\right)+1=1$ and $I\left(G\right)$ is principal, say $I\left(G\right)=\left(x_1{x}_2\right)$. In this case the thesis holds.

Suppose now $I(G\setminus \{1\left\}\right)\ne 0$. Then, by induction on n, $I(G\setminus \{1\left\}\right)$ is Cohen-Macaulay if and only if for all $2\le i\le j\le n$ such that $|N_{G\setminus \left\{1\right\}}\left(i\right)\cap \{i,\cdots ,n\}|,|N_{G\setminus \left\{1\right\}}\left(j\right)\cap \{j,\cdots ,n\}|>0$ and moreover for any $2\le i\le n$ such that $|N_{G\setminus \left\{1\right\}}\left(i\right)\cap \{i,\cdots ,n\}|>0$.

Notice that $N_G\left(i\right)\cap \{i,\cdots ,n\}=N_{G\setminus \left\{1\right\}}\left(i\right)\cap \{i,\cdots ,n\}$ for all $2\le i\le n$. Thus, combining (4) and (5) with the equality $pdS/I\left(G\right)=pdS/J=pdS/I(G\setminus \{1\left\}\right)+1$, we see that $I\left(G\right)$ is Cohen-Macaulay if and only if for all $2\le i\le n$ such that $|N_{G\setminus \left\{1\right\}}\left(i\right)\cap \{i,\cdots ,n\}|>0$.